SATHEE: Physics And Measurement Question 180

Physics And Measurement Question 180

Question: In an experiment, the following observation’s were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 $m / s^{2}$ using the formula , Y= $\frac{4 M g L}{π D^{2} l}$ , the maximum permissible error in Y is

Options:

A) 7.96%

B) 4.56%

C) 6.50%

D) 8.42%

Show Answer

Answer:

Correct Answer: C

Solution:

$Y = \frac{4 M g L}{π D^{2} l}$ so maximum permissible error in Y = $\frac{Δ Y}{Y} \times 100 = (\frac{Δ M}{M} + \frac{Δ g}{g} + \frac{Δ L}{L} + \frac{2 Δ D}{D} + \frac{Δ l}{l}) \times 100$

$= (\frac{1}{300} + \frac{1}{981} + \frac{1}{2820} + 2 \times \frac{1}{41} + \frac{1}{87}) \times 100$

$= 0.065 \times 100 = 6.5$ %

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Physics And Measurement Question 180

Question: In an experiment, the following observation’s were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 m/s2 using the formula , Y= 4MgLπD2l , the maximum permissible error in Y is

Options:

Answer:

Solution:

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Question: In an experiment, the following observation’s were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 $m / s^{2}$ using the formula , Y= $\frac{4 M g L}{π D^{2} l}$ , the maximum permissible error in Y is